Let \(f(z)=\left(-1-i\sqrt 3 \right)z+3-2i\). Describe \(f\) as a rotation followed by a dilation followed by a translation. Hence draw the image under \(f\) of the unit circle \(|z|=1\) and of the line \(x=y\).
Here, \(f(z)=2e^{-\frac{i2\pi}{3}}z+3-2i\). Thus \(f\) represents a clockwise rotation about the origin through angle \(\frac{2\pi}{3}\), followed by dilation with factor \(2\), followed by translation through \(3-2i\).
So, \(f\) takes the unit circle to the circle of radius \(2\) centred at \(3-2i\). We have \(-2\pi/3 + \pi/4 = -5\pi/12\), so \(f\) takes the line \(x=y\) to the line (with negative slope) meeting the real axis at angle \(-5 \pi/12\) passing through the point \(3-2i\).
Show that the function \(f(z)=az+b\) (with \(a\ne 0\)) may be described as a translation followed by a rotation followed by a dilation. Describe \(f(z)=\sqrt{2}\left(1-i\right)z+2-4i\) in this way.
We have \(az+b=a(z+(b/a))\), so \(f(z)=az+b\) can indeed be described as a translation through \(b/a\), followed by a rotation about the origin by angle \(\mathop{\text{Arg}}(a)\), followed by a dilation with factor \(|a|\).
For \(f(z)=\sqrt{2}\left(1-i\right)z+2-4i\) we have \[\begin{aligned} f(z)& \: = \: \sqrt{2}\left(1-i\right)z+2-4i\\ &= %2e^{-\frac{i\pi}{4}}z+2-4i %&=2\left( \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)\left(z+ \frac {2-4i}{\sqrt 2 (1-i)}\right) \\ %\: = \: \sqrt 2 (1-i) \left(z+\frac {2-4i}{\sqrt 2 (1-i)}\right) \: = \: 2e^{-\frac{i\pi}{4}}\left(z+\frac{3-i}{\sqrt{2}}\right). \end{aligned}\]
So \(f\) is translation through \(\frac{3-i}{\sqrt{2}}\), followed by rotation about the origin through an angle \(-\frac{\pi}{4}\) anticlockwise followed by dilation with factor \(2\).
In what subset of the complex plane is \(4z^3-3iz^2 + 4 - 3i\) conformal?
Let \(f(z)=4z^3-3iz^2 + 4 - 3i\); then \(f'(z)=12z^2-6iz=6z(2z-i)\). Thus for \(z\neq 0,i/2\) we have \(f'(z)\neq0\) and so \(f\) is conformal in \(\mathbb{C}\setminus\{0, i/2 \}\), but is not conformal at \(0\) or \(i/2\).
In what subset of the complex plane is \(2z^3-3\left(1+i\right)z^2+6iz\) conformal?
Let \(f(z)=2z^3-3\left(1+i\right)z^2+6iz\); then \[\begin{aligned} f'(z)&=6z^2-6\left(1+i\right)z+6i\\ &=6\left( z^2-\left(1+i\right)z+i\right) \\ &=6\left( z-1\right) \left(z-i\right)\ . \end{aligned}\] Thus for \(z\neq1,i\), we see that \(f'(z)\neq0\). Hence \(f\) is conformal in \(\mathbb{C}\setminus\{1,i\}\), but not for \(z\in\{1,i\}\) because \(f'(z)=0\) there.
In what subset of the complex plane is \(\sinh z\) conformal? For every point \(z_0\) at which the function is not conformal, give an example of two paths (lines) through \(z_0\) such that the angle (or the orientation of the angle) between them is not preserved by \(f(z)\) at \(z_0\).
Let \(f(z)=\sinh z\), so that \(f'(z)=\cosh z\). Certainly \(f(z)\) is holomorphic on \(\mathbb{C}\) and \[\begin{aligned} & f'(z)=0\\ \iff &\,e^z+e^{-z}=0\\ \iff &\,e^{2z}=-1\\ \iff &\,e^{2z}=e^{i\pi}\\ \iff & \,2z-i \pi = i2k\pi\\ \iff &\,{z}={i\left(\frac{\pi}{2}+k\pi\right)}\\ \end{aligned}\] where \(k\in\mathbb{Z}\). Let \[S=\left\{i\left(\frac{\pi}{2}+k\pi\right):k\in\mathbb{Z}\right\}\] Thus \(f(z)\) is conformal on \(\mathbb{C}\setminus S\), but not on \(S\).
Now take \(z_0=i\left(\frac{\pi}{2}+k\pi\right)\in S\). Using that \(\sinh(z+w)=\sinh z\cosh w+\cosh z \sinh w\) and \(\sinh(z_0)=i\sin(z_0/i)=i (-1)^k\), we get \(\sinh(z_0+w)=i (-1)^k\cosh w\). Consider the two lines \[\gamma_1(t)=z_0+t, \quad \gamma_2(t)=z_0+ i t\] for \(t\in\mathbb R\), then these curves meet at \(z_0\) at an angle \(\pi/2\). However, the curves \[(f\circ \gamma_1)(t)=i(-1)^k\cosh t, \quad (f\circ \gamma_2)(t)=i(-1)^k \cos t\] meet at \(f(z_0)\) at an angle \(\pi\). So the angle is not preserved at \(z_0\).
At which points in \(\mathbb C\) are
the following maps conformal?
(a) \(f(z)=z^3+2i\) (b) \(f(x+iy)=x-3yi\)
In both cases, for every point \(z_0\)
at which the function is not conformal, give an example of two paths
(lines) through \(z_0\) such that the
angle (or the orientation of the angle) between them is not preserved by
\(f(z)\) at \(z_0\).
(a) \(f(z)=z^3+2i\) is holomorphic, but \(f'(z)=3z^2\ne 0\) only when \(z\ne 0\). So \(f\) is conformal on \(\mathbb{C}^\ast = {\mathbb C}\setminus \{0\}\).
Any choices of two distinct lines passing through \(z=0\) works. For example, the paths \(\gamma_1(t)=(t-\tfrac{1}{2})e^{i\theta_1}\), \(\gamma_2(t)=(t-\tfrac{1}{2})e^{i\theta_2}\), \(t\in [0,1]\), \(\theta_1> \theta_2\) intersect at the origin at angle \(\theta_1-\theta_2\). However, \((f \circ \gamma_j)(t) = (t-\tfrac{1}{2})^3 e^{i3\theta_j} + 2i\) for \(j=1,2\). The angle between the images of these paths under \(f\) is therefore \(3\theta_1 - 3\theta_2\) (up to a multiple of \(2 \pi\)) [since \((f \circ \gamma_j)'(t) = 3(t-\tfrac{1}{2})^2 e^{i3\theta_j}\)], and so angle is not preserved.
(b) Let \(f(x+iy) = x-3yi\). We have \(u_x=1, v_y=-3, v_x=0, u_y=0\) so the first C-R equation is never satisfied. Therefore \(f\) is conformal nowhere.
For the second part of the question, we must check every point in the complex plane. Let \(z_0=x_0+iy_0\) be a complex number. Many choices of lines through \(z_0\) work, but let’s choose the simplest. Let \(\gamma_1(t) = z_0 + t\) and let \(\gamma_2(t) = z_0 + it\). The angle between \(\gamma_1\) and \(\gamma_2\) at \(z_0\) is clearly \(\pi/2\) anticlockwise (since \(\gamma_1'(t)=1\) and \(\gamma_2'(t)=i\)).
But, \((f \circ \gamma_1)(t)=(x_0+t)-3iy_0\), so \((f \circ \gamma_1)'(t)=1\), while \((f \circ \gamma_2)(t)=x_0-3i(y_0+t)\), so \((f \circ \gamma_2)'(t)=-3i\). The angle between \((f \circ \gamma_1)\) and \((f \circ \gamma_2)\) at \(z_0\) is therefore \(\pi/2\) clockwise. Hence orientation is not preserved.
Let \(f(z)=z^2+2z\). Show that \(f\) is conformal at \(z=i\) and describe the effect of \(f'(z)\) on the tangent vectors of curves passing through this point.
First note \[\begin{aligned} f'(z)&=2z+2\\ f'(i)&=2+2i\neq0 \end{aligned}\] so that \(f(z)\) is conformal at \(z=i\). Now, we have \(\mathop{\text{Arg}}( f'(i))=\dfrac{\pi}{4}\), so the tangent vectors at \(z=i\) are rotated by \(\dfrac{\pi}{4}\) anticlockwise. The dilation factor is \[\left|f'(i)\right|=\sqrt{4+4}=2\sqrt{2}.\]
Let \(f(z)=2z^3+3z^2\). Show that \(f\) is conformal at \(z=i\) and describe the effect of \(f'(z)\) on the tangent vectors of curves passing through this point.
First note \[\begin{aligned} f'(z)&=6z^2+6z\\ f'(i)&=-6+6i\neq0 \end{aligned}\] so that \(f(z)\) is conformal at \(z=i\). We have \(\mathop{\text{Arg}}(f'(i))=\dfrac{3\pi}{4}\), so the tangent vectors at \(z=i\) are rotated by \(\dfrac{3\pi}{4}\) anticlockwise. The dilation factor is \[\left|f'(i)\right|=\sqrt{6^2+6^2}=6\sqrt{2}.\]
Is the following true or false? If \(f,g\) are conformal at a point \(z_0\) then \(f+g\) is conformal at \(z_0\). Give a proof or a counter-example.
False. For example, let \(f(z)=z+z^2\), \(g(z)=-z\) and \(z_0=0\). Then \(f'(0)=1\), \(g'(0)=-1\) so both \(f\) and \(g\) are conformal at \(z=0\). But \(f(z)+g(z)=z^2\), which is not conformal at \(z=0\) because its derivative is 0 there.
Let \(f(z) = \overline{g(z)}\) with \(g(z)\) holomorphic (such functions \(f\) we call anti-holomorphic). Describe geometrically what happens to tangent vectors of paths passing through a point under the map \(f\). Conclude that \(f\) is angle-preserving, but reverses the orientation.
Let \(\gamma(t)\) be a path in \(\mathbb C\). Then \[\tfrac{\partial}{\partial t} f(\gamma(t)) \: =\: \tfrac{\partial}{\partial t} \overline{g(\gamma(t))} \: \stackrel{(\ast)}{=} \:\overline{ \tfrac{\partial}{\partial t} g(\gamma(t))} \:=\: \overline{g'(\gamma(t)) \cdot \gamma'(t)}.\] To see that the equality \((\ast)\) holds in the above, either observe that conjugation is preserved by linear maps, or explicitly notice: \[\left[\overline{(g \circ \gamma)(t)}\right]' \: = \: \left[u(\gamma(t)) - i v(\gamma(t))\right]' \: = \: \left[u(\gamma(t))\right]'-i\left[v(\gamma(t))\right]' = \overline{\left[u(\gamma(t))\right]'+i\left[v(\gamma(t))\right]'} = \overline{(g \circ \gamma)'(t)},\] where \(g=u+iv\).
Hence tangent vectors get rotated then dilated by the holomorphic derivative \(g'(z)\) of \(g\) as usual, but then they are conjugated (so reflected in the real axis). In the last step the angle is preserved, but orientation is reversed.
Let \(f: \mathcal D\to \mathcal D'\) be a biholomorphic map between two domains \(\mathcal D\) and \(\mathcal D'\). By considering the equation \(f (f^{-1}(w)) = w\) (for \(w \in \mathcal D'\)), show that \(f\) is conformal.
Since \(f\) is holomorphic with holomorphic inverse, we have that \(f \circ f^{-1}: \mathcal D' \to \mathcal D'\) is trivially holomorphic. Taking the derivative of both sides of the equation \(f (f^{-1}(w)) = w\), with respect to \(w\), it follows from the chain rule that \(f'(f^{-1}(w)) \cdot (f^{-1})'(w) = 1.\) Since \(f^{-1}\) is onto this means \(f'(z) \neq 0\) for all \(z \in \mathcal D\). Hence \(f\) is conformal on \(\mathcal D\).
Let \(\Omega:= \{z \in \mathbb C: \mathop{\text{Re}}(z) > \sqrt{3}\, |\mathop{\text{Im}}(z)| \}\). Sketch the domain \(\Omega\) and find its image \(f(\Omega)\) under the map \(f(z)=z^6\). Hence show that \(f\) is a biholomorphic map from \(\Omega\) onto its image, and give the inverse function.
The function is holomorphic on \(\mathbb C\). It is easy to see that \(\Omega= \{z \in \mathbb C: -\pi/6 < \mathop{\text{Arg}}(z) < \pi/6 \}\) (sketch of this sector is omitted). If \(z=|z|e^{i \mathop{\text{Arg}}(z)} \in \Omega\) then \(f(z) = |z|^6 e^{i 6\mathop{\text{Arg}}(z)}\). Since \(\mathop{\text{Arg}}(z) \in (-\pi/6, \pi/6)\) we have \(6\mathop{\text{Arg}}(z) \in (-\pi, \pi)\). Thus, the image of \(\Omega\) under \(f\) is \(\mathbb{C}\setminus \mathbb{R}_{\leq 0}\). By considering the modulus-argument form of any point in the image, it is clear that this map is injective and surjective. We know that the inverse map \(z^{1/6}= \exp(\frac 1 6 \mathop{\mathrm{Log}}(z))\) is holomorphic on \(\mathbb{C}\setminus \mathbb{R}_{\leq 0}\). Thus \[z^6: \Omega {\, \xrightarrow{\sim} \,}\mathbb{C}\setminus \mathbb{R}_{\leq 0}.\]